# Reciprocal of a vector

1. Dec 17, 2013

### Jhenrique

If:
$$\vec{v}=\frac{\mathrm{d} \vec{r}}{\mathrm{d} t}$$
so:
$$\\ \vec{v} = \frac{d\vec{r}}{dt} \\ \\ \vec{v}\;dt = d\vec{r} \\ \\ dt = \frac{d\vec{r}}{\vec{v}} \\ \\ \int dt = \int \frac{d\vec{r}}{\vec{v}} \\ \\ t = \int \frac{d\vec{r}}{\vec{v}}$$
Is true?
Solving the equation for time t, is need divide the position vector r by velocicty vector v... But I don't know do division between vectors...

2. Dec 17, 2013

### D H

Staff Emeritus
Division by a vector is not defined.

It appears you are trying to solve some problem and you thought this might be a way to attack that problem. What is the problem that are you trying to solve?

3. Dec 17, 2013

### Jhenrique

Given a vectorial velocity and a position vector, I want to calcule o time t.
In other words, I want solving the equation above for variable t.
$$\\ \vec{v} = \frac{d\vec{r}}{dt}$$

4. Dec 17, 2013

### HallsofIvy

Staff Emeritus
In general, if you are given $\vec{v}$ and $\frac{d\vec{r}}{dt}$, there will NOT BE a variable t such that $$\vec{v}= \frac{d\vec{r}}{dt}$$!

You can try $$\int \vec{v}dt= \int\vec{dr}$$ and then solve the resulting equation for t.

5. Dec 17, 2013

### ChrisVer

$\vec{r} \vec{u}= \vec{r} \frac{d\vec{r}}{dt}$
$\vec{r} \vec{u} =\frac{1}{2} \frac{dr^{2}}{dt}$
$dt= \frac{dr^{2}}{\vec{r} \vec{u}}=\frac{dr^{2}}{ru cosθ}$

is that correct?

6. Dec 18, 2013

### Jhenrique

In this case, the only solution is to consider the modulus of vectos: t = ∫ 1/v dr. Correct!?

7. Jan 5, 2014